We begin by noting we have been given no information about the time in which the stopping is to occur, only velocities and distance. This points us to the following kinematic equation:

\(\displaystyle V_f^2=V_i^2+2a\Delta x\)

By using the given information, and noting that coming to a stop implies a final velocity of \(\displaystyle 0 \frac{m}{s}\), we can directly substitute the numbers in and solve. It makes sense the final value is negative because the car is being accelerated in the negative direction in order to stop it.

\(\displaystyle a=\frac{-(36\frac{m}{s})^2}{2 (100m)}=-6.5 \frac{m}{s^2}\)

A tailwind is a wind traveling in the same direction as an object. The find the new speed of the airplane, we add the speed of the tailwind to the speed of the airplane.

\(\displaystyle \vec{v}= (37+9 )\frac{m}{s}=46\frac{m}{s}\)

Using the following kinematic equation:

 \(\displaystyle x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\) 

\(\displaystyle x\) is the final position, in meters, (\(\displaystyle 0m\) for ground).

\(\displaystyle x_{0}\) is the starting position (\(\displaystyle 20m\) above ground)

\(\displaystyle v_{0}\) is the starting velocity  \(\displaystyle (0\frac{m}{s})\) because the ball is dropped and therefor starts from rest).

\(\displaystyle a\) is the acceleration due to gravity given by the problem statement.

Input these values into the first ruling equation and note that in this notation up is the positive direction so the acceleration due to gravity is negative:

\(\displaystyle 0m=20m+(0\frac{m}{s})(t)+\frac{1}{2}(-10\frac{m}{s^{2}})t^{2}\)

Then we solve for \(\displaystyle t\).

\(\displaystyle 0=20+0-5t^{2}\)

\(\displaystyle -20=-5t^{2}\)

\(\displaystyle 4=t^{2}\)

\(\displaystyle t=2s\)

We use the equation:

\(\displaystyle x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\) 

With \(\displaystyle x_{0}=0m\), \(\displaystyle v_{0}=2\frac{m}{s}\), \(\displaystyle a=-10\frac{m}{s^{2}}\), and \(\displaystyle t=3s\) we have:

\(\displaystyle x=0+3*2+\frac{1}{2}*-10*3^{2}\)

\(\displaystyle x=-39m\)

The bag has fallen 39 meters from its original position. Note that the question does not ask for the distance the bag is from the basket after 3 seconds so we do not have to take in to account how far the balloon has moved in the 3 seconds. Also, note that the question asks "how far" the bag is from the basket, so we do not have to worry about sign in the answer.

We use the equation:

\(\displaystyle x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\) 

With \(\displaystyle x_{0}=0m\), \(\displaystyle v_{0}=200\frac{m}{s}\), \(\displaystyle a=-10\frac{m}{s^{2}}\) , and \(\displaystyle t=5s\) we have:

\(\displaystyle x=0+200*5+\frac{1}{2}*-10*5^{2}\)

\(\displaystyle x=875m\)

The part moves 875 meters upward in the time it breaks from the ship. Then, we use:

\(\displaystyle x=vt\) with \(\displaystyle v=200\frac{m}{s}\) and \(\displaystyle t=5s\) to find that the ship has moved 1000 meters during the same 5 second period.

So the ship moved upward 1000 meters in the 5 second time period while the part moved 875 meters upward so to find the distance between the ship and the part as the problem statement requested we subtract:

\(\displaystyle 1000m-875m=125m\)

We use the equation:

\(\displaystyle v_{f}^{2}=v_{0}^{2}+2ax\) 

Where the initial velocity is \(\displaystyle v_{o}=20\frac{m}{s}\), the acceleration is \(\displaystyle a=3\frac{m}{s^{2}}\) , and the displacement during the acceleration \(\displaystyle x=14m\).

Plug in those values into our equation and solve.

\(\displaystyle v_f^2=(20\frac{m}{s})^2+2*(3\frac{m}{s^2})*14m\)

\(\displaystyle v_f=\sqrt{400+84}=\sqrt{484}=22\frac{m}{s}\)

However unrealistic, if the acceleration of the race car were constant, we could calculate a value. We need everything in SI units, so we have to convert the distance to \(\displaystyle m\)

\(\displaystyle 1320ft\left(\frac{0.3048m}{1ft}\right )=402.3m\)

Using a kinematic equation, we can directly calculate the acceleration \(\displaystyle a\),

\(\displaystyle \Delta x = \frac{1}{2} a t^2 \Rightarrow a = \frac{2 \Delta x}{t^2}=\frac{2(402.3m)}{(3.5s)^2}=65.7\frac{m}{s^2}\)

To find the final velocity we use the following kinematic equation.

\(\displaystyle v_f^2=v_o^2+2a\Delta x\)

The block starts from rest, so \(\displaystyle v_o=0\). The incline is eighteen meters high, so we can find the distance (hypotenuse) as follows:

\(\displaystyle \Delta x=\frac{18m}{sin24^\circ}\)

\(\displaystyle \Delta x = 44.25m\)

We know that the acceleration is \(\displaystyle a=2.2\frac{m}{s^2}\).

Plug in the values to find the velocity at the bottom of the incline.

\(\displaystyle v_f^2=(0\frac{m}{s})^2+2(2.2\frac{m}{s^2})(44.25m)\)

\(\displaystyle v_f=\sqrt{194.7\frac{m^2}{s^2}}=14\frac{m}{s}\)

\(\displaystyle \overrightarrow{P}_{Total}=\overrightarrow{P}_1+\overrightarrow{P}_2...\)

Definition of momentum:

\(\displaystyle \overrightarrow{P}=m\overrightarrow{v}\)

Combine equations:

\(\displaystyle \overrightarrow{P}_{Total}=m_1\overrightarrow{v}_1+m_2\overrightarrow{v}_2\)

Plug in values:

\(\displaystyle \overrightarrow{P}_{Total}=25*< 10,0>+25*< 15,0>\)

\(\displaystyle \overrightarrow{P}_{Total}=< 250,0>+< 375,0>\)

\(\displaystyle \frac{\overrightarrow{P}_{Total}}{m}=\frac{< 625,0>kg\frac{m}{s}}{50kg}\)

\(\displaystyle \overrightarrow{v}=< 12.5,0>\frac{m}{s}\)

Let's first list the values we know.

\(\displaystyle a=10\frac{m}{s^2}\)

\(\displaystyle v_o=0\frac{m}{s}\)

\(\displaystyle d=45m\)

We are looking for the time it takes for this ball to reach the ground, thus we are looking for \(\displaystyle t\). The kinematics equation we need to use is:

\(\displaystyle d=v_ot +\frac{1}{2}at^2\)

Since the initial velocity is zero, the \(\displaystyle v_ot\) term disappears. Thus we are left with:

\(\displaystyle 45m=\frac{1}{2}*10\frac{m}{s^2}t^2\)

Solve for time.

\(\displaystyle t=3s\)

We only take the positive value in our calculation since time cannot be negative.